Extension Stoichiometry III Empirical Formula.pptx

IGCSE CHEMISTRY EMPIRICAL FORMULA NOTES

1. © 2004-05 Dorje Gurung Ext. Stoichiometry: Empirical Formula

2. Ext. Stoichiometry: Empirical Formula Slide 2 of 17 Learning objectives Concepts : subscript, empirical formula, empirical weight, molecular formula, structural formula Skills : Identify a given formula as empirical or molecular Given empirical formula and formula weight of the compound, determine the molecular formula

3. Ext. Stoichiometry: Empirical Formula Slide 3 of 17 Molecular and empirical formula Molecular Formula : formula that indicates the actual number (actual ratio) and types of atoms that make up the molecule (in the case of a covalent compound) or ions that make up the formula unit (in the case of an ionic compound) Empirical formula : gives only the relative number (smallest whole number ratio) of atoms

4. Ext. Stoichiometry: Empirical Formula Slide 4 of 17 Molecular and empirical formula Structural formula : formula that shows groups of atoms in a molecule that are bonded together. For example: CH 3 COOH which indicates that the first C is bonded to 3 H, the second C to 2 O (the molecular formula being C 2 H 4 O 2 ). CH 3 CH 2 CHOH which shows that the first C is bonded to 3 H’s the second C to 2 H’s, and the third C to 1 H and 1 O (the molecular formula being C 3 H 7 O). Structural formula becomes important when studying organic chemistry, and therefore we will return to it then.

5. Ext. Stoichiometry: Empirical Formula Slide 5 of 17 From Molecular to Empirical formula Molecular formula Empirical formula N 2 O 4 NO 2 P 4 O 6 P 2 O 3 C 2 H 4 CH 2 Mn 2 O 3 Mn 2 O 3 C 6 H 12 O 6 CH 2 O H 2 O H 2 O CO 2 CO 2 HNO 3 HNO 3 (NH 4 ) 2 SO 4 (NH 4 ) 2 SO 4 Fe 2 O 3 Fe 2 O 3

6. Ext. Stoichiometry: Empirical Formula Slide 6 of 17 From empirical to Molecular formula If all one knew about a compound was it’s empirical formula then the molecular formula cannot be determined. The form the molecular formula would take is the best one can do. Empirical formula form of molecular formula in the absence of other info NO 2 N x O 2x (could be one of NO 2 , N 2 O 4 , N 3 O 6 etc.) P 2 O 3 P 2x O 3x (could be one of P 2 O 3 , P 4 O 6 , P 6 O 9 etc.) CH 2 C x H 2x Mn 2 O 3 Mn 2x O 3x CH 2 O C x H 2x O x H 2 O H 2x O x CO 2 C x O 2x

7. Ext. Stoichiometry: Empirical Formula Slide 7 of 17 Determination of Empirical formula Since the subscripts in a formula (any formula) indicate the ratio in which the elements are combined to form the compound, determination of a formula is just a matter of determining that ratio. A formula such as N 2 O 4 tells us that a molecule of dinitrogen tetroxide contains 2 atoms of nitrogen and 4 of oxygen. Or, 1 mole of dinitrogen tetroxide contains 2 moles of nitrogen and 4 of oxygen atoms. Similarly, one mole of C 2 H 6 contains 2 moles of carbon and 6 of hydrogen atoms. So, the determination of empirical formula from available elemental analysis data involves the determination of the molar ratio or RAM ratio of the elements in the compound from their mass data.

8. Ext. Stoichiometry: Empirical Formula Slide 8 of 17 Determination of Empirical formula from elemental analysis I: Using the Mole a) From (weight/weight)% composition of the compound: Elemental analysis of a compound of mercury and chlorine showed the composition to be 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula of the compound? By using the Mole: Step 1 . Convert the percentage composition to mass ratio. If the composition is 73.9% mercury and 26.1% chlorine by mass, then a 100-gram sample of the compound would contain 73.9 g mercury and 26.1 g chlorine. Therefore, mass of Hg: mass of Cl = 73.9:26.1

9. Ext. Stoichiometry: Empirical Formula Slide 9 of 17 Determination of Empirical formula from elemental analysis I: Using the Mole Step 2 . Convert the mass ratio to mole ratio by using molar masses. Convert the masses into moles. # of moles of Cl = # of moles of Hg = So, the mole ratio of Hg to Cl = 0.368:0.735. (Hg 0.368 Cl 0.735 )

10. Ext. Stoichiometry: Empirical Formula Slide 10 of 17 Determination of Empirical formula from elemental analysis I: Using the Mole Step 3 . Find the number of moles of chlorine that combines with 1 mol of mercury. Mass of mercury/g # of moles of Mercury Mass of Chlorine/g # of moles of Chlorine 73.9 0.368 26.1 0.735  147.8 0.736 52.2 1.47 And, 295.6 1.47 104.4 2.94 Lastly, 201 1 71 2

11. Ext. Stoichiometry: Empirical Formula Slide 11 of 17 Determination of Empirical formula: Without Using the Mole a) From (weight/weight)% composition of the compound: Elemental analysis of a compound of mercury and chlorine showed the composition to be 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula of the compound? Without using the mole: Step 1 . Same as before.

12. Ext. Stoichiometry: Empirical Formula Slide 12 of 17 Determination of Empirical formula: Without Using the Mole Step 2 . Convert the mass ratio to ratio of RAMS atom ratio by using RAMs. # of RAMs of Cl = # of RAMs of Hg = So, the RAM ratio of Hg to Cl = 0.368:0.735. (Hg 0.368 Cl 0.735 )

13. Ext. Stoichiometry: Empirical Formula Slide 13 of 17 Determination of Empirical formula: Without Using the Mole Step 3 . Find the RAM of chlorine that combines with 1 RAM of mercury. Mass of mercury/g # of RAMs of Mercury Mass of Chlorine/g # of RAMs of Chlorine 73.9 0.368 26.1 0.735  147.8 0.736 52.2 1.47 And, 295.6 1.47 104.4 2.94 Lastly, 201 1 71 2